Abstract
This paper considers an observable double-ended queueing system of passengers and taxis, where matching times follow an exponential distribution. We assume that passengers are strategic and decide to join the queue only if their expected utility is nonnegative. We show that the strategy of passengers is represented by a unique vector of thresholds corresponding to different cases of the number of taxis observed in the system upon passenger arrival. Furthermore, we develop a heuristic algorithm to find an optimal range of fees to be levied on passengers to maximize social welfare or revenues.
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Acknowledgements
The research of Hung Q. Nguyen was supported by JST SPRING, Grant Number JPMJSP2124. The research of Tuan Phung-Duc was supported by JSPS KAKENHI Grant Numbers 18K18006 and 21K11765.
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Appendices
Appendix A Proof of Proposition 1
We will prove Proposition 1 by induction on p. The statement is equivalent to
for any fixed values of j.
Since \(T(0,j) = 0\), it is obviously implied from the recursive formulas that \(T(0,j) \le T(1,j)\); thus, (A1) holds with \(p=0\). Assuming that (A1) holds with \(p=q-1\) for any integer \(q\ge 1\), which indicates, for any fixed value of j,
We show that it holds with \(p=q\), which indicates that we need to prove that, for any fixed value of j,
by considering the following five cases.
-
When \(j=K\), from (2), we have
$$\begin{aligned} T(q,j)=\frac{1}{S\mu } + T(q-1,K-1), \end{aligned}$$(A3)and
$$\begin{aligned} T(q+1,j)=\frac{1}{S\mu } + T(q,K-1). \end{aligned}$$(A4)Since \(T(q-1,K-1) \le T(q,K-1)\) by assumption (A2), from (A3) and (A4), we obtain
$$\begin{aligned} T(q,j) \le T(q+1,j) \ \text {for} \ j=K. \end{aligned}$$(A5) -
When \(S<j<K\), from (2), we have
$$\begin{aligned} T(q,j) = \frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q,j+1) +\frac{S\mu }{\lambda _t+S\mu }T(q-1,j-1), \end{aligned}$$and
$$\begin{aligned} T(q+1,j) =\frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q+1,j+1) +\frac{S\mu }{\lambda _t+S\mu }T(q,j-1). \end{aligned}$$Now, due to (A5), it is seen that the inequality \(T(q,j) \le T(q+1,j)\) holds for \(j=K-1\) because
$$\begin{aligned} T(q,K-1)&= \frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q,K) +\frac{S\mu }{\lambda _t+S\mu }T(q-1,K-2)\\&\le \frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q+1,K)+\frac{S\mu }{\lambda _t+S\mu }T(q,K-2)\\&= T(q+1,K-1). \end{aligned}$$Then, it is easily obtained by induction on j, that
$$\begin{aligned} T(q,j) \le T(q+1,j) \ \text {for} \ S<j<K. \end{aligned}$$(A6) -
When \(j=0\), from (1), we have
$$\begin{aligned} T(q,0) =\frac{1}{\lambda _t} + T(q-1,1), \end{aligned}$$(A7)and
$$\begin{aligned} T(q+1,0) =\frac{1}{\lambda _t} + T(q,1). \end{aligned}$$(A8)Since \(T(q-1,1) \le T(q,1)\) because of the inductive assumption, from (A7) and (A8), we obtain
$$\begin{aligned} T(q,j) \le T(q+1,j) \ \text {for} \ j=0. \end{aligned}$$(A9) -
When \(0<j<S\), from (2), we have
$$\begin{aligned} T(q,j) =\frac{1}{\lambda _t+j\mu } +\frac{\lambda _t}{\lambda _t+j\mu }T(q-1,j+1) +\frac{j\mu }{\lambda _t+j\mu }T(q,j-1), \end{aligned}$$and
$$\begin{aligned} T(q+1,j) =\frac{1}{\lambda _t+j\mu } +\frac{\lambda _t}{\lambda _t+j\mu }T(q,j+1) +\frac{j\mu }{\lambda _t+j\mu }T(q+1,j-1), \end{aligned}$$Now, due to (A9), it is seen that the inequality \(T(q,j) \le T(q+1,j)\) holds for \(j=1\) because
$$\begin{aligned} T(q,1)&= \frac{1}{\lambda _t+\mu } +\frac{\lambda _t}{\lambda _t+\mu }T(q-1,2) +\frac{\mu }{\lambda _t+\mu }T(q,0) \\&\le \frac{1}{\lambda _t+\mu } +\frac{\lambda _t}{\lambda _t+\mu }T(q,2) +\frac{\mu }{\lambda _t+\mu }T(q+1,0) \\&= T(q+1,1). \end{aligned}$$Then, it is easily obtained by induction on j that
$$\begin{aligned} T(q,j) \le T(q+1,j) \ \text {for} \ 0<j<S. \end{aligned}$$(A10) -
Last, when \(j=S\), from (2), we have
$$\begin{aligned} T(q,j)=\frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q,S+1) +\frac{S\mu }{\lambda _t+S\mu }T(q,S-1), \end{aligned}$$(A11)and
$$\begin{aligned} T(q+1,j)=\frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q+1,S+1) +\frac{S\mu }{\lambda _t+S\mu }T(q+1,S-1). \end{aligned}$$(A12)However, note that \(T(q,S+1) \le T(q+1,S+1)\) and \(T(q,S-1) \le T(2,S-1)\) (implied from results (A6) and (A10)). Therefore, from (A11) and (A12), we obtain
$$\begin{aligned} T(q,j) \le T(q+1,j) \ \text {for} \ j=S. \end{aligned}$$(A13)
Equations (A5), (A6), (A9), (A10), (A13) complete our proof.
Appendix B Proof of Lemma 1
We prove Lemma 1 by induction on j.
First, note that
and
By induction on j, we have
for \(1\le \ j\le S-1\); and
for \(S+1\le \ j\le K-1\).
Finally,
It can also be noted that
and
Appendix C Proof of Lemma 2
We prove this lemma by induction on p. We can easily see that it holds with \(p=0\) and \(p=1\) due to Lemma 1. Assume that it holds with \(p=q-1\) for any integer \(q\ge 2\). Additionally, assume that when \(p=q-1\), the inequality holds in the case of \(j=S\). (We show that under the same inductive assumptions, it also holds when \(p=q\) and \(j=S\) later in Proposition 2.) Then, from assumptions, we have
We show that the inequality holds with \(p=q\), which indicates that we need to prove that
Assume there exists \(1\le j\le S-1\) such that
From (2), we have
which is equivalent to
On the other hand, we also have
according to the inductive assumption. From (C16), (C17) and (C18), we obtain
which implies
Additionally, from (2), we have
(due to (C19)). This implies
By induction on j, it finally implies
However,
which contradicts (C20). This indicates that (C15) holds and thus completes the proof.
Appendix D Proof of Lemma 3
We prove this lemma by induction on p. We can easily see that it holds with \(p=0\) and \(p=1\) due to Lemma 1. Assume that it holds with \(p=q-1\) for any integer \(q\ge 2\). Additionally, assume that when \(p=q-1\), the inequality holds in the case of \(j=S\). (We will show that, under the same inductive assumptions, it also holds when \(p=q\) and \(j=S\) later in Proposition 2.) Then, from assumptions, we have
We show that the inequality holds with \(p=q\), which indicates that we need to prove that
First, notice that
which indicates that (D22) holds with \(j=K\). Now, we make an inductive assumption on j; and for any \(S+2 \le j\le K-1\), consider the following:
Appendix E Proof of Proposition 2
We prove Proposition 2 by induction on p. The statement is equivalent to the following inequalities.
and
We already showed that (E23) and (E24) hold with \(p=0\) and \(p=1\) in Lemma 1. Assuming that (E23) and (E24) hold with \(p=q-1\) for any integer \(q\ge 2\), which indicates that
and
We show that it holds with \(p=q\), which indicates that we need to prove that
and
by considering the following five cases.
-
When \(j=0\), consider the following
$$\begin{aligned} T(q,1)-T(q,0)&= \left( \frac{1}{\lambda _t+\mu } +\frac{\lambda _t}{\lambda _t+\mu }T(q-1,2) +\frac{\mu }{\lambda _t+\mu }T(q,0) \right) - T(q,0) \\&= \frac{1}{\lambda _t+\mu } +\frac{\lambda _t}{\lambda _t+\mu }T(q-1,2) - \frac{\lambda _t}{\lambda _t+\mu }T(q,0) \\&= \frac{1}{\lambda _t+\mu } +\frac{\lambda _t}{\lambda _t+\mu }T(q-1,2) - \frac{\lambda _t}{\lambda _t+\mu }\left( \frac{1}{\lambda _t}+ T(q-1,1) \right) \\&= \frac{\lambda _t}{\lambda _t+\mu }\left( T(q-1,2)-T(q-1,1)\right) \\&\ge 0 \qquad \text {(due to the inductive assumption)}, \end{aligned}$$which indicates that
$$\begin{aligned} T(q,0) \le T(q,1). \end{aligned}$$(E25) -
When \(1\le j \le S-2\), from (2) we have
$$\begin{aligned} T(q,j) =\frac{1}{\lambda _t+j\mu } +\frac{\lambda _t}{\lambda _t+j\mu }T(q-1,j+1) +\frac{j\mu }{\lambda _t+j\mu }T(q,j-1), \end{aligned}$$(E26)and
$$\begin{aligned} T(q,j+1)&= \frac{1}{\lambda _t+(j+1)\mu } +\frac{\lambda _t}{\lambda _t+(j+1)\mu }T(q-1,j+2) \nonumber \\&\quad +\, \frac{(j+1)\mu }{\lambda _t+(j+1)\mu }T(q,j). \end{aligned}$$(E27)We prove \(T(q,j) \le T(q,j+1)\) by contradiction. Assuming \(\exists j,\ T(q,j) > T(q,j+1)\), combining with (E27), we have
$$\begin{aligned} T(q,j) > \frac{1}{\lambda _t+(j+1)\mu } +\frac{\lambda _t}{\lambda _t+(j+1)\mu }T(q-1,j+2) +\frac{(j+1)\mu }{\lambda _t+(j+1)\mu }T(q,j), \end{aligned}$$which is equivalent to
$$\begin{aligned} T(q,j) > \frac{1}{\lambda _t}+T(q-1,j+2). \end{aligned}$$However, we also have \(T(q-1,j+2) \ge T(q-1,j+1)\) (due to the inductive assumption), so
$$\begin{aligned} T(q,j) > \frac{1}{\lambda _t}+T(q-1,j+1). \end{aligned}$$(E28)From (E26) and (E28), we obtain
$$\begin{aligned} T(q,j)&< \frac{1}{\lambda _t+j\mu } +\frac{\lambda _t}{\lambda _t+j\mu }\left( T(q,j)-\frac{1}{\lambda _t}\right) +\frac{j\mu }{\lambda _t+j\mu }T(q,j-1), \end{aligned}$$which is equivalent to
$$\begin{aligned} T(q,j) < T(q,j-1). \end{aligned}$$By induction on j (by repeating the same procedure), it finally implies
$$\begin{aligned} T(q,1) < T(q,0), \end{aligned}$$which contradicts (E25) that is proved above. This contradiction indicates that
$$\begin{aligned} T(p,j) \ge T(p,j+1), \text {for}\ 1 \le j \le S-2. \end{aligned}$$(E29) -
When \(j=K-1\), consider the following
$$\begin{aligned}&T(q,K-1)-T(q,K) \\&\quad = \left( \frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q,K) +\frac{S\mu }{\lambda _t+S\mu }T(q-1,K-2) \right) - T(q,K) \\&\quad = \frac{1}{\lambda _t+S\mu } +\frac{S\mu }{\lambda _t+S\mu }T(q-1,K-2) - \frac{S\mu }{\lambda _t+S\mu }T(q,K) \\&\quad = \frac{1}{\lambda _t+\mu } +\frac{S\mu }{\lambda _t+S\mu }T(q-1,K-2) - \frac{S\mu }{\lambda _t+\mu }\left( \frac{1}{S\mu }+ T(q-1,K-1) \right) \\&\quad = \frac{S\mu }{\lambda _t+S\mu }\left( T(q-1,K-2)-T(q-1,K-1)\right) \\&\quad \ge 0 \qquad \text {(due to the inductive assumption)}, \end{aligned}$$which indicates that
$$\begin{aligned} T(q,K-1) \ge T(q,K). \end{aligned}$$(E30) -
When \(S+1\le j \le K-2\), from (2), we have
$$\begin{aligned} T(q,j) =\frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q,j+1) +\frac{S\mu }{\lambda _t+S\mu }T(q-1,j-1), \end{aligned}$$(E31)and
$$\begin{aligned} T(q,j+1) =\frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q,j+2) +\frac{S\mu }{\lambda _t+S\mu }T(q-1,j). \end{aligned}$$(E32)Due to the inductive assumption, we have \(T(q-1,j-1) \ge T(q-1,j)\); therefore, due to (E30), the inequality \(T(q,j) \ge T(q,j+1)\) holds for \(j=K-2\). By induction on j, we obtain
$$\begin{aligned} T(q,j) \ge T(q,j+1)\ \text {for}\ S+1 \le j \le K-2. \end{aligned}$$Next, we prove that \(T(q,S) \ge T(q,S+1)\). From (2), we have
$$\begin{aligned} T(q,S)=\frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q,S+1) +\frac{S\mu }{\lambda _t+S\mu }T(q,S-1), \end{aligned}$$(E33)and
$$\begin{aligned} T(q,S+1) = \frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q,S+2) +\frac{S\mu }{\lambda _t+S\mu }T(q-1,S). \end{aligned}$$To prove the above inequality, we show that
$$\begin{aligned} T(q,S-1)\ge T(q-1,S). \end{aligned}$$(E34)From (2), we have
$$\begin{aligned}&T(q,S-1) \\&\quad =\frac{1}{\lambda _t+(S-1)\mu } +\frac{\lambda _t}{\lambda _t+(S-1)\mu }T(q-1,S) +\frac{(S-1)\mu }{\lambda _t+(S-1)\mu }T(q,S-2) \\&\quad \ge \frac{1}{\lambda _t+(S-1)\mu } +\frac{\lambda _t}{\lambda _t+(S-1)\mu }T(q-1,S) \\&\qquad +\, \frac{(S-1)\mu }{\lambda _t+(S-1)\mu }\left( T(q,S-1) - \frac{1}{(S-1)\mu }\right) \\&\qquad \quad \text {(due to}\, \mathrm{Lemma\, 2}) \\&\quad = \frac{\lambda _t}{\lambda _t+(S-1)\mu }T(q-1,S) +\frac{(S-1)\mu }{\lambda _t+(S-1)\mu }T(q,S-1), \end{aligned}$$which implies that (E34) is true. Therefore,
$$\begin{aligned} T(q,S) \ge T(q,S+1). \end{aligned}$$(E35)$$\begin{aligned} T(q,S) \le \frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q,S) +\frac{S\mu }{\lambda _t+S\mu }T(q,S-1), \end{aligned}$$which implies
$$\begin{aligned} T(q,S-1) +\frac{1}{S\mu }\ge T(q,S), \end{aligned}$$and this also completes the proof of Lemma 2.
-
Finally, we prove that \(T(q,S) \ge T(q,S-1)\). To show the above equality, first note that
$$\begin{aligned}&T(q,S-1) \\&\quad =\frac{1}{\lambda _t+(S-1)\mu } +\frac{\lambda _t}{\lambda _t+(S-1)\mu }T(q-1,S) +\frac{(S-1)\mu }{\lambda _t+(S-1)\mu }T(q,S-2) \\&\quad \le \frac{1}{\lambda _t+(S-1)\mu } +\frac{\lambda _t}{\lambda _t+(S-1)\mu }T(q-1,S) +\frac{(S-1)\mu }{\lambda _t+(S-1)\mu }T(q,S-1), \end{aligned}$$due to (E29), and this implies
$$\begin{aligned} T(q,S-1) - T(q-1,S) \le \frac{1}{\lambda _t}. \end{aligned}$$(E36)Now, due to (E36), Lemma 3 and the inductive assumptions, we have
$$\begin{aligned}&T(q,S) - T(q,S+1) \\&\quad =\left( \frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q,S+1) +\frac{S\mu }{\lambda _t+S\mu }T(q,S-1) \right) \\&\qquad -\, \left( \frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q,S+2) +\frac{S\mu }{\lambda _t+S\mu }T(q-1,S) \right) \\&\quad = \frac{\lambda _t}{\lambda _t+S\mu }\left( T(q,S+1)-T(q,S+2)\right) +\frac{S\mu }{\lambda _t+S\mu }\left( T(q,S-1)-T(q-1,S)\right) \\&\quad \le \frac{\lambda _t}{\lambda _t+S\mu }.\frac{1}{\lambda _t} +\frac{S\mu }{\lambda _t+S\mu }.\frac{1}{\lambda _t}\\&\quad = \frac{1}{\lambda _t}, \end{aligned}$$which indicates that
$$\begin{aligned} T(q,S) \le \frac{1}{\lambda _t} + T(q,S+1). \end{aligned}$$(E37)(Note that this conclusion also completes the proof of Lemma 3).
Now, due to (E37), we have
$$\begin{aligned} T(q,S)&=\frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q,S+1) +\frac{S\mu }{\lambda _t+S\mu }T(q,S-1) \\&\ge \frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }\left( T(q,S) - \frac{1}{\lambda _t} \right) +\frac{S\mu }{\lambda _t+S\mu }T(q,S-1) \\&= \frac{\lambda _t}{\lambda _t+S\mu }T(q,S) +\frac{S\mu }{\lambda _t+S\mu }T(q,S-1), \end{aligned}$$which implies \(T(q,S) \ge T(q,S-1)\).
Appendix F Proof of Proposition 3
First, we prove that
for all \(p=1,2,\ldots .\)
This inequality holds for \(p=1\) because \(T(1,0)=\frac{1}{\lambda _t}\). Assume that it also holds for \(p=q \ge 1\), indicating that \(T(q,0) \ge \frac{q}{\lambda _t}\). We have
Therefore, by induction on p, we obtain that (F38) is true. \(\lim _{p \rightarrow +\infty } \frac{p}{\lambda _t} = +\infty \), which implies
By induction on j using formula (2), we can easily obtain
for all \(j=0,1,2,\ldots ,K\).
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Nguyen, H.Q., Phung-Duc, T. Strategic customer behavior and optimal policies in a passenger–taxi double-ended queueing system with multiple access points and nonzero matching times. Queueing Syst 102, 481–508 (2022). https://doi.org/10.1007/s11134-022-09786-3
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DOI: https://doi.org/10.1007/s11134-022-09786-3